So I'm having trouble understanding how to do initial value problems using y values with respect to x. For instance,
$\frac{dy}{dx} = 3x^{2}y^2$ where $y(1) = 1$
I know you can divide both sides by $y^2$ and then integrating on the RHS will give $x^3 + c$ but how does the LHS operate?
Also, in a situation where $\frac{dy}{dx} = 2 - 5y$ where $y(0) = 1$ you can rearrange it to get $\frac{dy}{dx} + 5y = 2$ But then there is apparently an integrating factor of $e^{5x}$ Where does that come from? What steps am I missing?